Convert Days to Year,Month,Day Format - php

Question

I have this problem. Can someone help me,how to convert number of days into the format XX Years, XX Month, XX Days...
i created this function,
function convert($sum) {
$years = ($sum / 365) ;
$years = floor($years);
$month = ($sum % 365) / 30.5;
$month = floor($month);
$days = ($sum % 365) % 30.5; // the rest of days
// Echo all information set
echo 'DAYS RECEIVE : '.$sum.' days<br>';
echo $years.' years - '.$month.' month - '.$days.' days';
}
convert(151);
But with 151 days the result was wrong
DAYS RECEIVE : 151 days0 years - 4 month - 1 days
it must be 4 month ans 28 days not 1 day...
http://sandbox.onlinephpfunctions.com/code/f5e6b4b4f6a27024b66ffbf04e80698722a3ecab

Answer 1

If you use more modern PHP, the following is based around actual days in each month:
$days = 151;
$start_date = new DateTime();
$end_date = (new $start_date)->add(new DateInterval("P{$days}D") );
$dd = date_diff($start_date,$end_date);
echo $dd->y." years ".$dd->m." months ".$dd->d." days";
Note that it will vary, depending on the current date, so you might prefer to set $start_date and $end_date to work from a fixed baseline
$days = 151;
$start_date = new DateTime('1970-01-01');
$end_date = (new DateTime('1970-01-01'))->add(new DateInterval("P{$days}D") );
$dd = date_diff($start_date,$end_date);
echo $dd->y." years ".$dd->m." months ".$dd->d." days";

Answer 2

This is the solution for your problem:
function convert($sum) {
$years = floor($sum / 365);
$months = floor(($sum - ($years * 365))/30.5);
$days = ($sum - ($years * 365) - ($months * 30.5));
echo “Days received: ” . $sum . “ days <br />”;
echo $years . “ years, “ . $months . “months, “ . $days . “days”;
}